Question 203724
{{{x^2 + 4y^2=4}}}
{{{(x-1)^2+y^2=1}}}
Normally one use the substituion method to solve this type of system. You would start by solving for one of the variables in one of the equations and proceed from there. And this system can be done that way, too. But this system, since Y appears only as a squared term, can be solved in other ways:<ul<li>We can solve for y^2 instead of y and substitute for y^2.</li><li>We can use the Elimination (aka Linear Combination or Addition) Method.</li></ul>
I will use the first one of these.<br>
Solve the second equation for y^2:
Subtract (x-1)^2 from both sides giving:
{{{y^2= 1 - (x-1)^2}}}
Substitute into the first equation for y^2:
{{{x^2 + 4(1 - (x-1)^2) = 4}}}
Solve for x. Start by simplifying:
{{{x^2 + 4(1 - (x^2 -2x + 1)) = 4}}}
{{{x^2 + 4(1 - x^2 + 2x - 1) = 4}}}
{{{x^2 + 4 - 4x^2 + 8x - 4) = 4}}}
{{{-3x^2 + 8x = 4}}}
Make one side 0:
{{{0 = 3x^2 - 8x + 4 = 0}}}
Factor: (or use the quadratic formula if you prefer)
{{{(3x - 2)(x - 2) = 0}}}
For this product to be zero one of the factors must be zero:
{{{3x - 2 = 0}}} or {{{x - 2 = 0}}}
{{{x = 2/3}}} or {{{x = 2}}}
Now we use these, one at a time, to find the y values which go with them:
For x = 2/3:
{{{(2/3)^2 + 4y^2 = 4}}}
{{{4/9 + 4y^2 = 4}}}
{{{4y^2 = 32/9}}}
{{{y^2 = 8/9}}}
{{{y = sqrt(8)/3 = (2sqrt(2))/3}}} or y = {{{-(2sqrt(2))/3}}}
For x = 2:
{{{(2)^2 + 4y^2 = 4}}}
{{{4 + 4y^2 = 4}}}
{{{4y^2 = 0}}}
{{{y = 0}}}<br>
So we have three points of intersection:
(2/3, {{{(2sqrt(2))/3)
(2/3, {{{-(2sqrt(2))/3)
(2, 0)